ncees exam 507 number of poles

[Sparky Bill PE]

Can someone try to put this in terms I can understand? The solution is you should just "assume 4 because the motor is going to run slightly faster than rated speed". 

I don't work in the motor fields, but if the motor is designed with 4 poles why wouldn't the name plate say it has 1,800 RPM? I'm having trouble trying to retain this so I don't forget and I don't want to treat it as a typical "NCEES tricky question" that don't make sense that they love to have. 

Like please don't explain it if you understand what NCEES is saying, I'm asking to explain it in a way that makes practical sense so I can learn from it and use it as an engineer. 

 

[akyip]

Can someone try to put this in terms I can understand? The solution is you should just "assume 4 because the motor is going to run slightly faster than rated speed". 

I don't work in the motor fields, but if the motor is designed with 4 poles why wouldn't the name plate say it has 1,800 RPM? I'm having trouble trying to retain this so I don't forget and I don't want to treat it as a typical "NCEES tricky question" that don't make sense that they love to have. 

Like please don't explain it if you understand what NCEES is saying, I'm asking to explain it in a way that makes practical sense so I can learn from it and use it as an engineer. 

View attachment 19654

Induction motors typically run at slightly lower than synchronous speed (actual speed n < ns). The given nameplate speed (1600 rpm) is the full-load actual speed n, so that means the synchronous speed ns must be just slightly higher than 1600 rpm.

Synchronous speed ns = 120 x f / p

The number of poles in an induction machine is always an even integer (corresponding to a set of N and S poles).

For p = 2 poles: ns = 120 x 60 / 2 = 3600 rpm... This is too high.

For p = 4 poles: ns = 120 x 60 / 4 = 1800 rpm. This is just about right since it is only a little bit higher than the nameplate full load speed of 1600 rpm.

Hence, the answer is 4 poles (4.0 poles = Choice B).

If you keep on increasing the number of poles, then the synch speed would be too low (you would have n > ns, which is not the case for an induction motor running at rated motor values).

Hope this helps.

 

[akyip]

Also to answer your question about why the motor nameplate would not list 1800 rpm:

Induction motors require a small difference between actual speed n and synchronous speed ns to generate torque for the load. If an induction motor ran at synchronous speed (n = ns), the result is that the motor does not produce any load torque (a no-load condition).

This is why the motor nameplate does not list the synchronous speed (1800 rpm in this problem). Rather, the nameplate lists the full-load rated speed n, which is only a little lower than ns.

 

[Dothracki PE]

Damn, @akyip beat me to it. I was going to say that basically the problem gives you the actual speed. The formula below will allow you to find the number of poles given most of the information in the problem. However if you plug in the 1600 rpm, the number of poles results in 4.5, which is a trick answer. Typical slip values are around 5-10% so the synchronous speed is roughly 5-10% greater than the nameplate or actual speed.



Also biggest tidbit of information for this problem that would elimiate the trick answer is that a motor cannot have a fractional amount of poles.

 

[akyip]

Damn, @akyip beat me to it. I was going to say that basically the problem gives you the actual speed. The formula below will allow you to find the number of poles given most of the information in the problem. However if you plug in the 1600 rpm, the number of poles results in 4.5, which is a trick answer. Typical slip values are around 5-10% so the synchronous speed is roughly 5-10% greater than the nameplate or actual speed.

View attachment 19655

LOL, my bad Dothracki.

Just trying to help. Trying to help other people on this board, to see how well I stack when I finally get to take my PE power exam.

Also, I'm going to emphasize once more that the number of poles p for an induction motor is always an even integer. This is because the number of N poles equals the number of S poles. With that said, you can easily eliminate Choice C (4.5 poles: this is not possible).

 

[DLD PE]

Can someone try to put this in terms I can understand? The solution is you should just "assume 4 because the motor is going to run slightly faster than rated speed". 

I don't work in the motor fields, but if the motor is designed with 4 poles why wouldn't the name plate say it has 1,800 RPM? I'm having trouble trying to retain this so I don't forget and I don't want to treat it as a typical "NCEES tricky question" that don't make sense that they love to have. 

Like please don't explain it if you understand what NCEES is saying, I'm asking to explain it in a way that makes practical sense so I can learn from it and use it as an engineer. 

View attachment 19654

They beat me to it too!  lol, but I will add:

You can never have half a pole, so just assign # poles at a time (2, 4, 6, etc.) until you get something that closely matches the nameplate rating.

n=120*f/p

2 poles:  n=120*60/2 = 3600 RPM (way off)

4 poles:  n=120*60/4 = 1800 RPM (close enough)

6 poles:  n=120*60/6 = 1200 RPM (4 poles is closer)

 

[Dothracki PE]

LOL, my bad Dothracki.

Just trying to help. Trying to help other people on this board, to see how well I stack when I finally get to take my PE power exam.

Also, I'm going to emphasize once more that the number of poles p for an induction motor is always an even integer. This is because the number of N poles equals the number of S poles. With that said, you can easily eliminate Choice C (4.5 poles: this is not possible).

Was also about to add that last point. No worries, as long as we all are giving correct feedback, it's good teamwork. Motors are not my strong suit although not my worst subject on the PE.

 

[Sparky Bill PE]

They beat me to it too!  lol, but I will add:

You can never have half a pole, so just assign # poles at a time (2, 4, 6, etc.) until you get something that closely matches the nameplate rating.

n=120*f/p

2 poles:  n=120*60/2 = 3600 RPM (way off)

4 poles:  n=120*60/4 = 1800 RPM (close enough)

6 poles:  n=120*60/6 = 1200 RPM (4 poles is closer)

Yeah this is the method I took. I knew it couldn't be 4.5. 

But then when i calculated 4 poles it came out to be 1800 RPM. I can't wrap my mind around the fact a motor will have more RPM's than its nameplate and that be okay. That is the Part I don't understand. I almost picked 4 because its "closer" but it didn't make sense, so I picked 6. Because the RPM rating is LESS than nameplate rating. Seems if a motor is going faster than nameplate that would be dangerous and not realistic. 

 

[akyip]

Also one more thing I forgot to add.

Slip s is calculated as:

s = (ns - n) / ns

It is basically the difference between synch speed ns and actual rotor speed n, divided by synch speed ns. You can also think of it as the difference in speed expressed as a percentage of synch speed ns.

As Dothracki mentioned, the slip s is typically a small percentage value. I've never seen a slip s over 10%. Hence, actual speed n should not be too much below synch speed ns.

Using the above slip formula, you can calculate the actual rotor speed as:

n = ns (1 - s)

Likewise, you can calculate synch speed as:

ns = n / (1 - s)

 

[Dothracki PE]

Yeah this is the method I took. I knew it couldn't be 4.5. 

But then when i calculated 4 poles it came out to be 1800 RPM. I can't wrap my mind around the fact a motor will have more RPM's than its nameplate and that be okay. That is the Part I don't understand. I almost picked 4 because its "closer" but it didn't make sense, so I picked 6. Because the RPM rating is LESS than nameplate rating. Seems if a motor is going faster than nameplate that would be dangerous and not realistic. 

It's because the rotor and the magnetic field that causes the rotation rotate at different speeds, typically the rotor speed is slower than the field rotation for an induction motor. The rotor is what puts out the mechanical power so that's typically the nameplate or actual speed.

 

[Sparky Bill PE]

It's because the rotor and the stator rotate at different speeds, typically the rotor slower than the stator. The rotor is what puts out the mechanical power so that's typically the nameplate or actual speed.

Someone correct me if that's wrong.

Are you implying the nameplate speed is for the stator speed and the actual speed is the rotor speed? 

I still don't get it, because by this methodology you could buy a motor with a nameplate speed of 1,800 and the damn thing run at 18,000 RPM. 

 

[Dothracki PE]

Are you implying the nameplate speed is for the stator speed and the actual speed is the rotor speed? 

I still don't get it, because by this methodology you could buy a motor with a nameplate speed of 1,800 and the damn thing run at 18,000 RPM. 

Sorry that was not correct. I didn't finish reading my reference before I replied. 

The synchronous speed is the speed of the magnetic field rotation that is induced by the stator. The magnetic field causes the rotor to rotate. The stator is a stationary component. 

Synchronous motors operate at synchronous speed. Whereas induction motors (like the one in the problem) have a 5-10% difference (also known as slip) between synchronous and actual speed. 

 

[Sparky Bill PE]

Sorry that was not correct. I didn't finish reading my reference before I replied. 

The synchronous speed is the speed of the magnetic field rotation that is induced by the stator. The magnetic field causes the rotor to rotate. The stator is a stationary component. 

Synchronous motors operate at synchronous speed. Whereas induction motors (like the one in the problem) have a 5-10% difference (also known as slip) between synchronous and actual speed. 

That makes perfect sense, but that speed difference is based of the slip. This problem only gives a scenario as if you were handed a nameplate without slip, just speed. So you have to take the RPM at face value and guess at how many poles it has. I don't know why anyone would guess that the number of poles it has makes the actual machine "faster" (higher RPMs). 

Thanks for writing it out I agree with what you just said and remembered that the entire time working this problem. 

 

[speakeelsy PE]

Are you implying the nameplate speed is for the stator speed and the actual speed is the rotor speed? 

I still don't get it, because by this methodology you could buy a motor with a nameplate speed of 1,800 and the damn thing run at 18,000 RPM. 

The basic idea is that sync speed is faster than rotor speed. The nameplate always list the shaft speed for that motor - which is always going to be less than sync speed because of all the losses (friction, windage, etc.)

nameplate RPMs are less than synchronous speeds because your motor induction motor will never turn exactly at synchronous speed -- i've honestly never seen one as low as the one in the problem. usually its around 1765 for a 4 pole motor.

The magnetic field in the stator rotates at sync speed always (this depends on the frequency of your power source - 60Hz here in the state) (Stators STAY put, meaning they don't move)

In and induction motor, the magnetic field created by the stator induces the rotor poles to follow the stator magnetic field and ROTATE. Since they have moving parts connected to them, like a load, and bearings, and such, they are susceptible to losses.

Check out Wildi's and Chapman's books induction motor chapters, and maybe youtube a motor dissection video?

 

[jd5191]

I really like this video for an explanation of slip and why rotor is slightly slower than sync speed on an induction motor

 

[wfg42438]

Induction motors typically run at slightly lower than synchronous speed (actual speed n < ns). The given nameplate speed (1600 rpm) is the full-load actual speed n, so that means the synchronous speed ns must be just slightly higher than 1600 rpm.

Synchronous speed ns = 120 x f / p

The number of poles in an induction machine is always an even integer (corresponding to a set of N and S poles).

For p = 2 poles: ns = 120 x 60 / 2 = 3600 rpm... This is too high.

For p = 4 poles: ns = 120 x 60 / 4 = 1800 rpm. This is just about right since it is only a little bit higher than the nameplate full load speed of 1600 rpm.

Hence, the answer is 4 poles (4.0 poles = Choice B).

If you keep on increasing the number of poles, then the synch speed would be too low (you would have n > ns, which is not the case for an induction motor running at rated motor values).

Hope this helps.

Thanks for this clear explanation!!!
I was having the same issue as @Sparky Bill PE